# I Why $\alpha=atan(v/c)$?

#### PeterDonis

Mentor
I meant 4-vctirs by $\vec{v_1}$ and $\vec{v_2}$.
Ok, then what do you think their 4-vector dot product means, physically? Or, for that matter, what do you think each 4-vector by itself means, physically? What do these 4-vectors represent?

#### olgerm

Gold Member
Now I am quite sure, that it is $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$, because otherwise would be that
$\alpha_{t-t'}=arccosh(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccosh(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccosh(\frac{1}{\sqrt{1-v^2/c^2}})$
, but I was told that $\alpha_{t-t'}=arccos(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccos(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$.

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#### PeterDonis

Mentor
Now I am quite sure, that it is $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$,
If they are 4-vectors, you are wrong. You haven't answered the questions I asked about what these vectors mean, physically. If you don't know the answer to that you aren't going to understand anything else.

Can you answer the question, what do the 4-vectors $\vec{v_1}$ and $\vec{v_2}$ mean physically? Or don't you know? If you don't know, then you need to figure that out first before even trying to guess about what their dot product and its relationship with the rapidity $\alpha$ might be.

#### olgerm

Gold Member
If they are 4-vectors, you are wrong. You haven't answered the questions I asked about what these vectors mean, physically. If you don't know the answer to that you aren't going to understand anything else.

Can you answer the question, what do the 4-vectors $\vec{v_1}$ and $\vec{v_2}$ mean physically? Or don't you know? If you don't know, then you need to figure that out first before even trying to guess about what their dot product and its relationship with the rapidity $\alpha$ might be.
These are quantities that describe location in SR-spacetime. I am not sure what kind of answer you espect.

#### PeterDonis

Mentor
These are quantities that describe location in SR-spacetime.
No, they aren't. We are not dealing with spacetime position vectors here--if we were, everything that has been written in this thread would be nonsense.

I am not sure what kind of answer you espect.
In other words, you don't know.

The standard math of SR describes the worldline of a particle with a 4-velocity vector, usually denoted $u$, which is a unit vector tangent to the worldline at a particular event. If we have two particles in relative motion, we can choose the event to be the point in spacetime where they pass each other, and then we have two 4-velocity vectors $u_1$ and $u_2$. Their dot product is $u_1 \cdot u_2 = \gamma$, where $\gamma$ is the relativistic gamma factor corresponding to the ordinary "relative velocity" between the particles, i.e., $\gamma = 1 / \sqrt{1 - v^2}$.

In terms of rapidity, we have $\gamma = \cosh \alpha$, and also $v = \tanh \alpha$. From this it is easy to calculate that $\gamma v = \sinh \alpha$. So we have $u_1 \cdot u_2 = \cosh \alpha$. Note that, since $u_1$ and $u_2$ are both unit vectors, we could indeed write the dot product as $u_1 \cdot u_2 = |u_1| |u_2| \cosh \alpha$, since both magnitudes are $1$. But $\alpha$ here, as has been said, is the rapidity; it is not an "angle in spacetime between the worldlines" in any useful sense, since it is not limited to the range $0$ to $2 \pi$.

#### PeterDonis

Mentor
But $\alpha$ here, as has been said, is the rapidity; it is not an "angle in spacetime between the worldlines" in any useful sense, since it is not limited to the range $0$ to $2 \pi$.
And just to repeat once more, this means that the rapidity $\alpha$ is not the angle $\alpha$ that appears in the Wikipedia article linked to in the OP. The angle that appears in the Wikipedia article has no useful physical meaning.

#### Ibix

I don't think rapidity can be correctly called an "angle". It certainly is not an "angle between axes in spacetime" in the sense the OP seems to want to interpret that; rapidities are not limited to the range $0$ to $2 \pi$. See my previous post in response to the OP.
Fair point. But I think it is fair to say that rapidity is analogous to angle in the same way that interval is analogous to distance. Like the angle in Euclidean geometry, rapidity is a measure of how far off parallel two lines are and is additive. But it can take any value, not just $0-2\pi$, fundamentally because it is not periodic. As rapidity, I think it's also only usually defined for timelike vectors since they are the only things that correspond to something moving. It's certainly possible to write the inner product of a timelike and a spacelike vector, but if you try to derive a rapidity from it you'll get a complex number. Not necessarily a bad thing, but maybe goes some way to explaining why it doesn't get used so much.

Anyway, since there are all these differences you cannot easily read the rapidity off a Minkowski diagram.

#### olgerm

Gold Member
I meant that
4-vectors indicate direction and magnitude in spacetime(by direction I do not mean spatial direction, but direction in spacetime that includes time).
Spacetime position 4-vectors are quantities that describe displacement from coordinate 0-point in spacetime(by displacement I do not mean spatial displacement, but displacement in spacetime that includes time).

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#### olgerm

Gold Member
it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$. Then $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$

then:
$\frac{e_t'}{i}\cdot e_x'=|e_t'/i|*|e_x'|*cos(\alpha_{t'-x'})$
$e_t\cdot e_x=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$0=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$\alpha_{t-x}=arccos(0)$
$\alpha_{t-x}=\frac{2\pi}{4}$

#### vanhees71

Gold Member
Should I use $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cosh(\alpha)$ instead of $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$?
No! No! No! You should use the correct math. You are completely mislead by whatever book. A Minkowski plane is not a euclidean plane! For a simple introduction, see

#### Nugatory

Mentor
it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$.
No. Introducing that factor of $i$ allows many equations to take on the same form as they do for Euclidean space, but the similarity is superficial and just in the mathematical formalism. The space is still non-Euclidean in ways that cannot be avoided - for example, a straight line is not, in general, the shortest distance between two points.

There's some history here. My old copy of Goldstein and many other textbooks of that vintage used the $ict$ formalism. However, by the mid-1970's MTW had a short section stating that our old friend $ict$ was to be "put to the sword", something that had to be unlearned.

• Ibix and vanhees71

#### Ibix

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$. Then $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$

then:
$\frac{e_t'}{i}\cdot e_x'=|e_t'/i|*|e_x'|*cos(\alpha_{t'-x'})$
$e_t\cdot e_x=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$0=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$\alpha_{t-x}=arccos(0)$
$\alpha_{t-x}=\frac{2\pi}{4}$
As @Nugatory says, not really. It's a mistake to try in my opinion. One of the key points for clear thinking is to make assumptions and differences between concepts as clear as possible. The $ict$ approach has always struck me as attempting to bury something quite subtle, and it comes back to bite when you move on to GR.

Honestly, I think the best approach is to explain that the inner product of two vectors, usually written $\vec v^T\vec v$ (in matrix notation, assuming $\vec v$ is a column vector), is more generally written as $\vec v^T\mathbf g\vec v$, where $\mathbf g$ is the metric tensor, which can be written as a square matrix. To get Euclidean geometry in Cartesian coordinates you set $\mathbf g$ to the identity matrix (which is why you don't normally see it written). To get Minkowski geometry in Einstein coordinates you make one of the diagonal elements negative (all of special relativity and our notions of cause and effect spring from that one change!). To get general relativity you let it get more complicated still.

This way you can see that the rule for taking inner products is the same in Minkowski and Euclidean spaces - but the spaces are fundamentally different in ways encoded in the metric tensor. And you are laying the groundwork for general relativity at the same time. My 2p, anyway.

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#### vanhees71

Gold Member
To make the concepts even clearer, rather call $\mathbf{g}$ the components of the fundamental form, since in general it's not a metric, because it's not positive definite (but nondegenerate). Sometimes one calls it a pseudometric in this case. It's simply a non-degenerate bilinear form on a vector space.

#### olgerm

Gold Member
So on euclidean papersheet:
$\alpha_{t-t'}=arctan(v/c)$
$\alpha_{t-x}=\frac{2\pi}{4}$
$\alpha_{t-x'}=\frac{2\pi}{4}-arctan(v/c)$
$\alpha_{t'-x}=\frac{2\pi}{4}-arctan(v/c)$
$\alpha_{t'-x'}=\frac{2\pi}{4}-2*arctan(v/c)$
$\alpha_{x-x'}=arctan(v/c)$

And really in spacetime:
$\alpha_{t-t'}=arccos(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccos(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$
$\alpha_{t-x}=arccos(-\vec{e_x}\cdot\vec{e_t})=\frac{2\pi}{4}$
$\alpha_{t-x'}=arccos(-\vec{e_x'}\cdot\vec{e_t})=arccos(-(\gamma∗\beta∗\vec{e_t}+\gamma∗\vec{e_x})* \vec{e_t})=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x}=arccos(-\vec{e_x}\cdot\vec{e_t'})=arccos(-\vec{e_x}\cdot(\gamma∗\vec{e_t}+\beta∗\gamma∗\vec{e_x}))=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x'}arccos(-\vec{e_x'}\cdot\vec{e_t'})=\frac{2\pi}{4}$
$\alpha_{x-x'}=arccos(\frac{\vec{e_x}\cdot \vec{e_x'}}{|\vec{e_x}|*|\vec{e_x'}|})= arccos(\frac{\vec{e_x}\cdot (\beta*\gamma*\vec{e_t}+\gamma*\vec{e_x})}{(1)*(1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$

Can you confirm these are correct?

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#### PeterDonis

Mentor
Can you confirm these are correct?
The "Euclidean papersheet" values look fine, but have no physical meaning.

The "really in spacetime" values are all nonsense except for the first one, if the first one is interpreted as the rapidity. Besides the rapidity $\alpha$, which you appear to be calling $\alpha_{t-t'}$, the only other meaningful quantities expressible in terms of trig functions (hyperbolic or otherwise) are $\gamma = \cosh (v / c)$ and $\gamma v = \sinh (v / c)$; but neither of those appear on your list.

• vanhees71 and Pencilvester

#### olgerm

Gold Member
I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.

#### PeterDonis

Mentor
I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.
That doesn't change anything I said in post #65.

At this point you have been given the correct information multiple times. There is no point in simply continuing to repeat the same corrective information if you aren't going to accept it.

• vanhees71

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